Problem: Rewrite the function by completing the square. $f(x)= 2x^{2}+3x-2$ $f(x)=$
Answer: $\begin{aligned} f(x)&=2x^2+3x-2 \\\\ &=2\left(x^2+\dfrac32x\right)-2 \end{aligned}$ Now we want to complete $x^2+\dfrac32x$ into a perfect square. To do that, we should add $\left(\dfrac{{+\frac32}}{2}\right)^2={\dfrac{9}{16}}$ to it: $x^2{+\dfrac32}x+{\dfrac{9}{16}}=\left(x+\dfrac34\right)^2$ We add ${\dfrac{9}{16}}$ inside the parentheses, and subtract ${2}\cdot{\dfrac{9}{16}}$ outside them, to keep the expression equivalent. $\begin{aligned} &\phantom{=}2\left(x^2+\dfrac32x\right)-2 \\\\ &=2\left(x^2+\dfrac32x+{\dfrac{9}{16}}\right)-2-2\cdot{\dfrac{9}{16}} \\\\ &=2\left(x+\dfrac34\right)^2-2-\dfrac98 \\\\ &=2\left(x+\dfrac34\right)^2-\dfrac{25}{8} \end{aligned}$ In conclusion, the function after completing the square is written as: $f(x)=2\left(x+\dfrac34\right)^2-\dfrac{25}{8}$